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Xenon (xe) of mass 5.08 g reacts with fluorine to form 9.49 g of a xenon fluoride compound. what is the empirical formula of this compound?

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User Thorvald
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1 Answer

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We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.

mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g

Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F

The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
XeF₆
answered
User Aslam Patel
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7.9k points
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